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Let $f: \mathbb{R} \to \mathbb{R}$ of class $C^2$ i.e $f$ is differentiable with continuous second derivative. Suppose that $f''(x)>a$ for all $x\in \mathbb{R}$ and some $a>0$. Proof that $f$ has a absolute minimun.

I have tried to solve this problem without any success, my attempt is:

Let $x>0$. By fundamental theorem of calculus we have that $\int_{0}^{x} f''(t)dt=f'(x)-f'(0).$ Then $$f'(x)=f'(0)+\int_{0}^{x} f''(t)dt\geq f'(0)+ax$$ for all $x>0$.

On the other hand, note that $\int_{0}^{x} f'(t)dt=f(x)-f(0)$ so $$f(x)=f(0)+\int_{0}^{x} f'(t)dt\geq f(0)+\int_{0}^{x} (f'(0)+at)dt$$ I just conclude

$$f(x)\geq f(0)+f'(0)x+\frac{1}{2}ax^2 \text{ for all } x\in \mathbb{R}$$

I have only found that $f$ is above a quadratic function but I would actually think that the local minimum would be the vertex of the quadratic function.

Any suggestion is appreciated.

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  • $\begingroup$ In your solution the continuity of $f''$ is required. The crucial inequality follows from the MacLaurin formula for $n=2,$ where $f$ is twice differentiable. $\endgroup$ Feb 23 at 17:07

2 Answers 2

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Your inequality shows that the function $f$ tends to $+\infty$ both for $x$ going to $-\infty$ and to $+\infty$.

A continuous function that has $+\infty$ as its limit both for $x$ going to $-\infty$ and to $+\infty$ has a minimum.

This follows from the fact that there is a closed interval $[-A,A]$ outside of which $f$ is larger than $f(0)$. Then use the fact that a continuous function on $[-A,A]$ has always a minimum.

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Just observe that $f'(x) $ is strictly increasing and hence it either tends to a limit or to $\infty$ as $x\to \infty$. If it tends to a limit $L$ then $$f'(x+1)-f'(x)\to L-L=0$$ By mean value theorem the left hand side equals $f''(c) $ and it thus always exceeds a positive number $a$. So the above equation can't hold and thus $f' (x) \to\infty $ as $x\to\infty $.

Similarly $f'(x) \to-\infty $ as $x\to-\infty $. And by intermediate value theorem we see that $f'$ vanishes somewhere say at $c$. Since the derivative $f'$ is strictly increasing, $f'$ vanishes only at $c$. Then $f'<0$ in $(-\infty, c) $ and $f'>0$ in $(c, \infty) $. Then $f$ is strictly decreasing in $(-\infty, c] $ and strictly increasing in $[c, \infty) $. Then $f$ attains absolute minimum at $c$.

There is no need to assume continuity of second derivative.

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